Solution: i
Math 132 Spring 2013 Final Exam
1.
Let
F( x ) =
x 2 ln( 1 + t2 + t3 )
ln( 1 + t )
dt.
Calculate
F' (4), the derivative of
F(x)
at
x = 4.
0
a) 0 b) 1 c) 2 d) 3 e) 4 f) 5 g) 6 h) 7 i) 8 j) 9
Solution: i
> F := (x) -> Int(2*ln(1+t^2+t^3)/ln(1+sqrt(t)), t = 0 .. x);
F
:=
x
x 2
ln( 1 + t2 ln( 1 +
+ t3 ) t)
dt
0
> D(F)(x);
#
# This is calculated using the Fundamental Theorem of Calculus
- without any integration
2 ln( 1 + x2 + x3 )
ln( 1 + x ) > derivative := D(F)(4);
Answer = simplify( derivative );
2 ln( 81 ) derivative :=
ln( 1 + 4 )
Answer = 8
2. Calculate e 12 1
1 + 8 ln( x ) dx.
x
a) 21
b) 22 c) 23 d) 24 e) 25
f) 26
g) 27 h) 28 i) 29 j) 30
Solution: f
> J := Int(12*sqrt(1+8*ln(x))/x,x = 1 .. exp(1)); # # Given integral
J := e 12
1 + 8 ln( x ) dx
x
1
> K := changevar(u = 1 + 8*ln(x), J, u);
#
# Integral with same value obtained by substitution u =
1+8*ln(x), du = (1/x)*dx
> K = value(K);
K := 9 3 2 u du
1
9 3 2 u du = 26
1
Verification using Maple's built-in integrator
> J = int(12*sqrt(1+8*ln(x))/x,x = 1 .. exp(1));
e 12
1
1 + 8 ln( x ) dx = 26
x
3.
Calculate
2
4 x(
x x
+ +
3 1)
dx
.
1
a) ln( 2 ) f) ln( 9 )
b) ln( 3 ) c) ln( 4 ) d) ln( 6 ) e) ln( 8 ) g) ln( 12 ) h) ln( 16 ) i) ln( 18 ) j) ln( 24 )
Solution: g
> with(student): > J := Int((4*x+3)/x/(x+1),x = 1 .. 2);
J
:=
2
4 x(
x x
+ +
3 1)
dx
1
> processedIntegrand := convert( integrand(J), parfrac, x);
#
# Partial fraction form of integrand
31 processedIntegrand := +
x x+1 > antiderivative := int(processedIntegrand, x);
# # Indefinite integral of given definite integral
antiderivative := 3 ln( x ) + ln( x + 1 ) > answer := simplify(subs(x = 2, antiderivative) - subs(x = 1,
antiderivative)); # # Value of given integral
answer := 2 ln( 2 ) + ln( 3 ) > Displayed_answer = combine(answer, ln);
# # Combine logarithms to get listed answer: Uses p*ln(u) = ln(u^p) and ln(x)+ln(y) = ln(x*y)
ln( 12 )
Verification using Maple's built-in integrator:
> Int((4*x+3)/x/(x+1),x = 1 .. 2) = int((4*x+3)/x/(x+1),x = 1 .. 2);
2
4 x(
x x
+ +
3 1)
dx
=
2
ln( 2 )
+
ln( 3 )
1
> testeq( 2*ln(2)+ln(3) = ln(12) );
true
4. Calculate
1
4 x2 ( x2
+ +
x 1
+ 4 2 )
dx.
0
1 a) +
4 f) 2 + 1
1 b) +
2
g) + 1
4
c) + 1
h) + 2
4
1 d) 2 +
4
i) + 1
2
1 e) 2 +
2
j) + 2
2
Solution: a
> J := Int( (4*x^2+x+4)/(x^2+1)^2, x = 0 .. 1);
J
:=
1
4 x2 + x + 4
(
x2
+
1
2
)
dx
0
> processedIntegrand := convert( student[integrand](J), parfrac,
x);
#
# Partial fraction form of given integrand
x
4
processedIntegrand :=
+
(
x2
+
1
2
)
x2 + 1
> antiderivative := int(processedIntegrand, x);
#
# Antiderivative for integrand of given integral
1
antiderivative := -
+ 4 arctan( x )
2 ( x2 + 1 )
> Answer := simplify(subs(x = 1, antiderivative) - subs(x = 0,
antiderivative));
1 Answer := +
4
Verification using Maple's built-in integrator:
> J = value(J);
1
4 x2 ( x2
+ +
x+4
2
1)
dx
=
1 4
+
0
5.
Given that
2
5
(2 e
x
)
cos( x )
dx
=
e
-
2,
what is
2
5
( e
2
x
)
sin(
x
)
dx
?
0
0
a) e - 3
b) e - 1
c) e + 1
d) e + 2
e) e + 3
f) 2 e - 3
g) 2 e - 1 h) 2 e + 1
i) 2 e + 2 j) 2 e + 3
Solution: h
> J := Int(5*exp(2*x)*sin(x),x=0..Pi/2); # # J is the definite integral to be evaluated
J
:=
2
5
(2
e
x
)
sin(
x
)
dx
0
> K := intparts(J,exp(2*x));
#
# Applying integration by parts to J with u = exp(2*x) and dv =
5*sin(x)*dx
# The expression K has the same value as J
................
................
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